Java中的模式编码[封闭]

Question

4

bbb*bbb
bb*i*bb
b*iii*b

import java.util.*;

public class Main {

    public static void main(String[] args) {

        int n, m, j, k, l, p, o = 0;
        Scanner s = new Scanner(System.in);
        n = s.nextInt();
        for (m = 1; m <= n; m++) {
            for (j = m; j < n; j++) {
                System.out.print("b");
            }

            for (k = 2; k < 2 * m + 1; k++) {
                System.out.print("*");
            }
            for (o = 3; o < m + 2; o++) {
                System.out.print("i");
            }
            for (p = 3; p < m + 2; p++) {
                System.out.print("*");
            }
            for (l = m; l < n; l++) {
                System.out.print("b");
            }
            System.out.println();

        }
    }
}

bbbb*bbbb
bbb***i*bbb
bb*****ii**bb
b*******iii***b
*********iiii****

Answer 1

所以

• 多了一行
• *总是一个，不需要循环
• i应使用row*2作为绑定

java prettyprint-override">for (row = 0; row < n - 1; row++) {
    for (j = row; j < n - 1; j++) {
        System.out.print("b");
    }
    // always
    System.out.print("*");
    for (i = 1; i < row * 2; i++) {
        System.out.print("i");
    }
    if (row != 0) { // except first line
        System.out.print("*");
    }
    for (k = row; k < n - 1; k++) {
        System.out.print("b");
    }
    System.out.println();
}

使用字符串。repeat

for (row = 0; row < n - 1; row++) {
    String b = "b".repeat(n - 1 - row);
    String i = "i".repeat(Math.max(0, row * 2 - 1));
    System.out.println(b + "*" + i + (row != 0 ? "*" : "") + b);
}